Newtons Third Law. How is jumping from a pier different to jumping from a boat? This is the absolute basic of a physics and yet 2 hours of googling fails to find an answer ! So ignoring all vertical movement and just concentrating on the horizontal movement:-

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*A man who weighs 75 kg jumps off a pier (steps off horizontally) with a force of 150N. Ignoring gravity he accelerates by 2 meters per sec per sec.


*A man who weighs 75kg jumps off (steps off horizontally) a boat that weighs 75 kg with a force of 150N. Ignoring gravity he accelerates by 2 meters per sec per sec. So does the boat (it accelerates by same amount), but in the opposite direction.


*What is wrong here? Im sure the boat and the man would end up at half the speed in each direction compared to example 1. Example 2 sounds like free energy compared to example one. What am I missing, apart from a working version of google, and an easy way to format this text?


*p.s.  I think one of my key confusions is in example 2,  is the acceleration relative to a point in space or is the acceleration relative to the other body. I.e. do they accelerate away from each other at 2m per sec^2 or 4m per sec^2? I guess my other confusion is if they are accelerating away at twice the rate then I appear to be doing twice the amount of work for the same amount of force applied which seems odd, and feels like free energy.
 A: Good question, though you have stated it a little imprecisely I think. So first I will state the question more precisely.
Scenario 1. Man leaps from pier, such that the force between himself and the ground, in the horizontal direction, reaches 150 N (for some length of time to be discussed).
Scenario 2. Man leaps from a boat having the same mass as the man, such that the force between himself and the boat, in the horizontal direction, reaches 150 N (for some length of time to be discussed).
You are correct to suppose that in both cases the acceleration of the man, relative to some convenient frame of reference such as planet Earth, reaches the same value in the two scenarios at the moment when the force reaches 150 N. But in the second scenario the man will find it harder to get the force to reach that value, because the boat is being pushed away as his legs get straighter. He could do it, but he would find that he had to expend more energy.
To find the final velocity of the man in the two cases, you can use either energy or momentum. In terms of momentum, what you need to know is the length of time for which any give force is applied. Leaping from a pier, a man can get to some given force pretty quickly, before his body has moved very much, and then maintain that force for, let's say, half a second. So he gets the momentum approximately 75 kg m/s. Leaping from a boat, the force will rise from zero more slowly, so will take longer to get to 150 N. So he will only manage to apply 150 N for some shorter time. His final momentum will therefore be lower than in first case.
Next let's think about energy. Now we have to consider the distance over which the centre of mass of the man moved while the force applied to the centre of mass was 150 N. Leaping from the pier, this distance is approximately the distance by which his legs extended from the crouch to the leap. Leaping from the boat, since the boat moves away, his centre of mass moves by about the half the distance of the first case. Therefore the energy delivered to the man in the second case is about half that of the first case. On this estimate the man does the same work in the two cases, but when leaping from the boat half of the work goes to kinetic energy of the boat, and half to the man, whereas in the first case the man gets all the kinetic energy (the change in motion of the pier being negligible).
A: To avoid getting sidetracked into questions of how muscles work, let's replace the man with a sack of potatoes with a mass of $75$ kg and let's suppose the force is provided by a compressed spring which produces an initial force of $150$ N when it is released.
If the sack of potatoes is on a pier (or on the ground) then the spring exerts a force of $150$ N on the sack and (by Newton's third law) on the pier. The man has a mass of $75$ kg and so initially accelerates at $2$ metres per second per second. The pier has (effectively) the mass of the whole Earth, so its acceleration is negligible. If the sack travels a small distance $d$ (small enough so that the force exerted by the spring does not change) the work done by the spring is $150d$ Joules. This energy was originally potentially energy in the compressed spring, and becomes the kinetic energy of the sack. Note that the pier does not move (or, to be precise, its movement is negligible) so the spring does no work on the pier, and the kinetic energy of the pier is negligible.
If the sack of potatoes is on a boat (or on a trolley) which has a mass of $75$ kg then the spring exerts a force of $150$ N on the sack and on the boat. Again, the man has a mass of $75$ kg and so initially accelerates at $2$ metres per second per second. But this time the boat has a mass of $75$ kg so its acceleration is $2$ metres per second per second in the opposite direction. This time, when the sack travels a small distance $d$ (small enough so that the force exerted by the spring does not change) then the boat also travels the same distance $d$ in the opposite direction, so the total work done by the spring is now $300d$ Joules. So both the sack and the boat accelerate and gain kinetic energy, but the work done by the spring is twice what it was in the first case.
A: As @anna_v has addressed several times in the comments, your confusion is caused by acceleration not being a conserved quantity. That it, the sum of accelerations is not necessary the same at different times. The total momentum and energy is, however, conserved at all times. As a small exercise, it might be fruitful for you to compute the total momentum of the system and check that it vanishes at all times.
A: In the first case the man does a force in the pier and the pier does a force in the man. Both of 150 N. Of course only during the leg's impulse.
In the second case, the man does a force in the boat and the boat does a force in the man. Both of 150 N.
No difference concerning the third law.
And third law is conservation of momentum:
$$F_{12} = -F_{21} => \frac{\partial \mathbf p_2}{\partial t} = -\frac{\partial \mathbf p_1}{\partial t} $$
About conservation of momentum, the second case is easy because the objects have the same mass. In the first one, the pier is attached to the Earth, so all the Earth is accelerated backwards by the force. But as its mass is so big, the acceleration is negligible.
